A machine in an ice factory exerts 3.06 ✕ 102 N of force to pull a large block of ice up a frictionless slope at constant speed. The force is applied parallel to the slope, and the weight of the block is 1.39 ✕ 104 N. What is the angle of the slope?
mgsinα=F
sinα=F/mg=3.06x10^6/1.39x10^4 =0.022
α= 1,26 degr
To find the angle of the slope, we need to use the concept of forces and trigonometry.
First, let's break down the forces involved:
1. The weight of the block of ice: This force acts vertically downward and has a magnitude of 1.39 ✕ 104 N.
2. The force applied by the machine: This force acts parallel to the slope and has a magnitude of 3.06 ✕ 102 N.
Since the block is being pulled up the slope at a constant speed, the force applied by the machine is equal in magnitude and opposite in direction to the component of the weight of the block along the slope. In other words, these forces are balanced.
Now, let's calculate the component of the weight of the block along the slope:
The component of the weight of the block along the slope can be determined using trigonometry. It is given by the formula:
Component of weight along slope = Weight of block × sin(θ)
where θ is the angle of the slope.
Rearranging the formula, we have:
sin(θ) = (Component of weight along slope) / (Weight of block)
Substituting the given values:
sin(θ) = (3.06 ✕ 102 N) / (1.39 ✕ 104 N)
Now, let's solve for θ:
θ = sin^(-1) [(3.06 ✕ 102 N) / (1.39 ✕ 104 N)]
Using a calculator, we can find the arcsine of the ratio to get the angle θ.
Please note that the value obtained will be in radians. If you want to convert it to degrees, multiply the result by (180/π).